Transcript Molar Volume

Molar Volume
Molar Volume
• There is a way to relate the volume
of a gas to the amount of gas in
moles.
• The volume of a gas is related to
both its temperature and its pressure
• This must be considered to
understand the relationship between
volume and moles
Temperature and Pressure
• Both affect the volume of gases
• Example-Balloon in freezer
• If temperature decreases, volume of
gas decreases
• Example-Diver in deep ocean
• If pressure is increased, volume of
gas is decreased
Standard temp and
press
• Standard conditions
-The average pressure of the atmosphere
at sea level is taken as standard pressure
which is 101.3kPa
-The freezing point of water is defined as
standard temperature, which is 0°C or
273°K
• Together these conditions are known as
standard temperature and pressure or
STP
Avogadro’s Hypothesis
• Used the findings of Guy-Lussac and Dalton
• Avogadro said: equal volumes of all ideal gases
at the same temp. and press. contain the same
number of molecules
Where n=# of
• Mathematically:
moles, V=volume,
nαV or n=kV or n1/V1=n2/V2 and k=a constant
• Based on the law, one mole of gas occupies the
same volume as one mole of another gas at the
same temp and pressure
• Molar volume of a gas us the space that is
occupied by one mole of the gas. Units=L/mol
Standard Molar Volume
• The molar volume of a gas at STP is
22.4L/mol
• Can use this number to solve gas
problems involving moles and
volume at STP
Example
• What is the volume of 3.0mol of
nitrous oxide, NO2(g), at STP?
Solution
• We know 1.00mol of gas occupies
22.4L at STP so this tells us that
n1=1.0mol and V1=22.4L
• The question tells us that the n2 is
3.0mol of NO2
• All we have to do is use the equation
(avogadro’s law)and solve for V2
Solution
• n1/V1=n2/V2
V2=n2V1
n1
V2=(3.0mol)(22.4L)
1.0mol
V2=67L
Example
• Suppose you have 44.8L of methane
gas at STP
a) How many moles are present?
b) What is the mass of the gas?(g)
c) How many molecules of gas are
present?
Solution
a) Using Avogadro’s law:
n2=n1V2
V1
=1.00mol x 44.8L
22.4L
=2.00mol